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7nadin3 [17]
2 years ago
12

1. The solubility of AgNO3 at 20°C is 222.0g AgNO3/100g H2O. What mass of AgNO3 can be dissolved in 250 g of water at 20°C? Reca

ll that solubility = mass of solute/ mass of solvent.
2. The solubility of methane, the major component of natural gas, in water at 20°C and 1.00 atm pressure is 0.026 g/L. If the temperature remains constant, what will be the solubility of this gas at 0.06 atm pressure? Recall the relationship between solubility and pressure for gases: S1/P1 = S2/P2
Chemistry
1 answer:
neonofarm [45]2 years ago
5 0

Answer :

(1) The mass of silver nitrate is, 555 g

(2) The solubility of the gas will be, 0.433 g/L

<u>Solution for Part 1 :</u>

From the given data we conclude that

In 100 gram of water, the amount of silver nitrate = 222 g

In 250 gram of water, the amount of silver nitrate = \frac{222}{100}\times 250=555g

Therefore, the mass of silver nitrate is, 555 g

<u>Solution for Part 2 :</u>

Formula used : S_1P_1=S_2P_2     (at constant temperature)

where,

S_1 = initial solubility of methane gas = 0.026 g/L

S_2 = final solubility of methane gas

P_1 = initial pressure of methane gas = 1 atm

P_2 = final pressure of methane gas = 0.06 atm

Now put all the given values in the above formula, we get the solubility of methane gas.

(0.026g/L)\times (1atm)=S_2\times (0.06atm)

S_2=0.433g/L

Therefore, the solubility of the gas will be, 0.433 g/L

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3 years ago
An analytical chemist weighs out 0.093g of an unknown monoprotic acid into a 250mL volumetric flask and dilutes to the mark with
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Answer:

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Explanation:

<u>Step 1:</u> The balanced equation

HA(aq) + NaOH(aq) → NaA(aq) + H2O(l)

It takes 1 mole of NaOH to neutralize 1 mole of the triprotic acid. This is called the reaction stoichiometry.

<u>Step 2:</u> Data given

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titrates with 0.16 M NaOH

adds 6.5 mL NaOH

<u>Step 3: </u>Calculate moles of NaOH

We know the concentration and volume of NaOH needed to neutralize the acid.

By determining the moles of NaOH in that volume in liters (95.9mL=0.0959L), the moles of acid in the original sample can be determined from the reaction stoichiometry.

Moles = Molarity * Volume

Moles = 0.16 M * 0.0065 L

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<u>Step 4: </u>Calculate moles of the unknown acid:

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For 0.00104 moles NaOH we have 0.00104 moles of HA

<u>Step 5: </u>Calculate the molar mass of the acid

Molar mass Ha = Mass Ha / moles Ha

Molar mass Ha = 0.093 grams / 0.00104 moles

Molar mass Ha = 89.42 g/mol ≈89 g/mol

The molar mass of the unknown acid is 89 g/mol

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